Let f(x, y) and g(x, y) ∈ ℝ[x, y] be two polynomials of degree 2. Assume that there exists ε ∈ ℝ such that Z_{fg+ε}(ℝ) is a set of four points. Then it can be shown that there exists α(x, y) ∈ ℂ[x, y] and c ∈ ℝ such that

fg + ε = cα\overline{α}.

Let us only check here that if fg + ε factors in ℂ[x, y], then it factors as above. Suppose fg + ε = dαβ with deg(α) = 1. Then deg(β) = 3 and Z_{fg+ε}(ℝ) cannot have only four points. Assume now that deg(α) = deg(β) = 2. Without loss of generality we may assume that both α and β have a coefficient equal to 1. Write dαβ = d\overline{α}\overline{β}. Since the factorization is unique, we must have α = a\overline{α} or α = a\overline{β} for some a ∈ ℂ. The case α = a\overline{α} implies that a = 1 and α = \overline{α}. Then β = \overline{β} and the curve Z_{fg+ε}(ℝ) cannot be a set of four points since a conic has either no points over ℝ or infinitely many. Hence, we have α = a\overline{β} for some a ∈ ℂ. Thus β = \overline{α}/\overline{a} and we find that fg + ε = (d/\overline{a})α\overline{α}. It follows that c := d/\overline{a} ∈ ℝ.

Note that if P ∈ Z_{fg+ε}(ℝ), then P ∈ Z_{α}(ℝ)∩Z\overline{α}(ℝ). To find a factorization of fg + ε explicitly, use the following geometric fact:

Fact: Take two conics h(x, y) and m(x, y) that both contain a given set of four points. Then any other conic that passes through the four points has the form h + λm with λ ∈ ℂ.

In our case, we can choose h and m to be in ℝ[x, y], since we can take for h and m the product of equations of lines passing through the (real) points in Z_{fg+ε}(ℝ). Hence, having chosen h and m ∈ ℝ[x, y], we need to solve for λ in

fg + ε = c(h + λm)(h + \overline{λ}m)
= c(h^{2} + |λ|m^{2} + (λ + \overline{λ})hm).